Normal modes of vibration from the total energy

[Jaime_mc2]

A mass $m$ is restricted to move in the parabola $y=ax^2$, with $a>0$. Another mass $M$ is hanging from this first mass using a spring with constant $k$ and natural lenghth $l_0$. The spring is restricted to be in vertical position always. The coordinates for the system are $x$ (horizontal coordinate of mass $m$) and $y$ (vertical coordinate of mass $M$).

With this system in mind, I was told to look for the equilibrium points using Lagrangian formulation, then approximate the kinetic and potential energy for small oscillations around that point, and finally, obtain the small oscillation frequency and normal modes of vibration from the total energy.

The first part was quite easy. I found the Lagrangian $$ L = \dfrac{1}{2}m(1+4a^2x^2)\dot{x}^2 + \dfrac{1}{2}M(\dot{x}^2+\dot{y}^2) -mgax^2 - Mgy - \dfrac{1}{2}k(ax^2-y-l_0)^2\ , $$ and applying the Euler-Lagrange equations I found the equations of motion, from where I could calculate the equilibrium point $(x_0,y_0) = (0,y_0)$.

Now it is where the problems start to appear. By using a Taylor expansion around the equilibrium point, I found the approximations for the kinetic and potential energy, which are $$ T \approx \dfrac{1}{2}(m+M)\dot{x}^2 + \dfrac{1}{2}M\dot{y}^2 \quad\text{y}\quad V \approx a\left[mg-k(y_0+l_0)\right]x^2 + \dfrac{1}{2}k(y-y_0)^2\ , $$ where I removed the constant term $V(0,y_0)$ for simplicity.

I don't understand how I am suppose to get the normal modes of oscillation from the total energy. I thought about applying energy conservation but, apparently, this is only recommended when the system is one-dimensional. Since the Lagrangian does not depend on time, neither the coordinates and potential, I assumed that what they wanted is to associate the total energy with the Hamiltonian, so I computed it: $$ H \approx \dfrac{p_x^2}{2(m+M)} + \dfrac{p_y^2}{2M} + a\left[mg-k(y_0+l_0)\right]x^2 + \dfrac{1}{2}k(y-y_0)^2 $$ Applying the cannonical equations, what I get is the system $$ \left\{\begin{array}{l} (m+M)\ddot{x} + 2a\left[mg-k(y_0+l_0)\right]x = 0 \\ M\ddot{y} + k(y-y_0) = 0 \end{array}\right.\ , $$ which is not coupled, and $x$ and $y$ seem to oscillate in different independent frequencies.

How am I supposed to get normal modes from a system where the variables oscillate independently? Also, is there any other way in which I can use the total energy to solve the problem?

 

[Paul Colby]

How am I supposed to get normal modes from a system where the variables oscillate independently?

I can't see anything wrong with your development so, if correct, aren't the modes are already normal?

Okay, what happen to the constraint $y=ax^2$?

 

[Paul Colby]

Okay, I think $y$ is the contact point with the parabola. One must also account for the length of the spring with an additional variable, say $z$. $x$ determines $y$ through the constraint.

 

[etotheipi]

Okay, I think $y$ is the contact point with the parabola. One must also account for the length of the spring with an additional variable, say $z$. $x$ determines $y$ through the constraint.

I think you are right, the mistake is even in the problem statement:

A mass $m$ is restricted to move in the parabola $y=ax^2$, with $a>0$. Another mass $M$ is hanging from this first mass using a spring with constant $k$ and natural lenghth $l_0$. The spring is restricted to be in vertical position always. The coordinates for the system are $x$ (horizontal coordinate of mass $m$) and $y$ (vertical coordinate of mass $M$).

when really the vertical coordinate of $M$ is not $y$, but rather $y - z = ax^2 - z$. That changes a few parts of the Lagrangian!